Solve 21/1+i

$$\frac { 21 } { 1 + i }$$

Evaluate

$$\frac{21}{2}-\frac{21}{2}i=10.5-10.5i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

$$\frac{21\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{21\left(1-i\right)}{1^{2}-i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{21\left(1-i\right)}{2}$$

Multiply 21 times 1-i.

$$\frac{21\times 1+21\left(-i\right)}{2}$$

Do the multiplications in 21\times 1+21\left(-i\right).

$$\frac{21-21i}{2}$$

Divide 21-21i by 2 to get \frac{21}{2}-\frac{21}{2}i.

$$\frac{21}{2}-\frac{21}{2}i$$

Real Part

$$\frac{21}{2} = 10\frac{1}{2} = 10.5$$

Solution Steps

Multiply both numerator and denominator of \frac{21}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(\frac{21\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{21\left(1-i\right)}{1^{2}-i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{21\left(1-i\right)}{2})$$

Multiply 21 times 1-i.

$$Re(\frac{21\times 1+21\left(-i\right)}{2})$$

Do the multiplications in 21\times 1+21\left(-i\right).

$$Re(\frac{21-21i}{2})$$

Divide 21-21i by 2 to get \frac{21}{2}-\frac{21}{2}i.

$$Re(\frac{21}{2}-\frac{21}{2}i)$$

The real part of \frac{21}{2}-\frac{21}{2}i is \frac{21}{2}.

$$\frac{21}{2}$$